1
$\begingroup$

Let $U$ be an open subset of $\mathbb{R}^m$ and $f:U\rightarrow \mathbb{R}^n$ a function. Fix $a\in U$.

Suppose $v$ is any non-zero vector in $\mathbb{R}^m$ and that $$L=\lim_{t\rightarrow 0} \frac{f(a+tv)-f(a)}{t}$$ exists and is independent of $v$ (i.e. it is same for all non-zero vectors $v$).

So directional derivative of $f$ at $a$ in all directions exists and is same.

Q. Can we conclude that $f$ is differentiable at $a$? If not, is it continuous at $a$?


The examples in books are given about existence of directional derivative in each direction but function is not continuous such as $f(x,y)=\frac{xy^2}{x^2+y^4}$ for $x\neq 0$ and $f(0,y)=0$. In this example, directional derivative of $f$ at $0$ exists along all directions; but its values are different. If $v=(a_1,a_2)$ then directional derivative is $a_2^2/a_1$ for $a_1\neq 0$, and its $0$ if $a_1=0$. So values of directional derivatives are different.

For the existence of derivative, we not only take linear directions to reach at $a$, but all curvilinear directions. I didn't find this question (mentioned above) in the books, stated as exercise at least, where above examples is mentioned.

$\endgroup$
1
$\begingroup$

Define $f:\mathbb R^2\to \mathbb R$ by setting $f=1$ on the half parabola $\{(x,x^2): x>0 \},$ and $f=0$ everywhere else. Then all directional derivatives of $f$ at $(0,0)$ are $0.$ But $\lim_{x\to 0^+}f(x,x^2)=1\ne0= f(0,0).$ Thus $f$ is not even continuous at $(0,0),$ much less differentiable there.

$\endgroup$
1
$\begingroup$

If $L=L(v)$ is independent of $v$, then $L\equiv 0$, since from the definition of $L$ we obtain that $L(-v)=-L(v)$.

Set $$ f(x,y)=\left\{\begin{array}{ccc} \dfrac{xe^{-1/y^2}}{x^2+e^{-2/y^2}} & \text{if} & y\ne 0 \\ 0 & \text{if} & y=0. \end{array} \right. $$ Then for all $b\ne 0$, $$ \frac{f(ta,tb)-f(0,0)}{t}=\dfrac{ae^{-1/(t^2b^2)}}{t^2a^2+e^{-2/(t^2b^2)}}= \dfrac{a}{t^2a^2e^{1/(t^2b^2)}+e^{-1/(t^2b^2)}}\to 0 $$ For $b=0$, the limit above is also zero. But $f$ is not even continuous at $(0,0)$.

Consider $$ \big(x(t),y(t)\big)=(ce^{-1/t^2},t)\to (0,0), \quad \text{as $t\to0$} $$ and $$ f\big(x(t),y(t)\big)=\frac{ce^{-2/t^2}}{(c^2+1)e^{-2/t^2}}=\frac{c}{c^2+1}\in [-1/2,1/2]. $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.